Related rate of volume change in a draining conical tank

Since the water forms a cone similar to the tank, the radius and depth satisfy $\frac{r}{h} = \frac{3}{10}$, so $r = \frac{3}{10}h$. Substituting into $V = \frac{1}{3}\pi r^2 h$ gives $V = \frac{1}{3}\pi \left(\frac{3}{10}h\right)^2 h = \frac{3\pi}{100}h^3$, expressing volume purely in terms of the single variable $h$.

Differentiating both sides with respect to $t$ using the chain rule gives $\frac{dV}{dt} = \frac{9\pi}{100}h^2 \frac{dh}{dt}$. Substituting $h = 4$ ft and $\frac{dh}{dt} = -\frac{1}{2}$ ft/min yields $\frac{dV}{dt} = \frac{9\pi}{100}(16)\left(-\frac{1}{2}\right) = -\frac{72\pi}{100} = -\frac{18\pi}{25}$ ft$^3$/min, so the volume is decreasing at about 2.26 ft$^3$/min.